2021 WSMO Team Round/Problem 4

Creating a triangle with vertices $A_1$ $B_1$ and $C_1$ , we immediately notice that triangle $A_1 B_1 C_1$ is a 30-60-90 triangle, with hypotenuse $A_1 C_1$ , and a longest leg $B_1 C_1$ .

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Next, we read the next part of the problem. We create another right triangle similar to the original, but with hypotenuse $B_1 C_1 = 3\sqrt{3}$ , and legs $B_1 B_2 = \frac{3\sqrt{3}}{2}$ and $C_1 B_2 = \frac{9}{2}$ .

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Since all the resulting triangles are similar, and we need to find the summation of the area of all of the resulting triangles (to infinity), we must use an infinite geometric series to determine the total area.

The area of the first triangle is $\frac{9\sqrt{3}}{2}$ , and the second triangle (that is also similar to the first triangle), $\frac{27\sqrt{3}}{8}$ .

Thus, since all the resulting triangles are similar to the one before it, the common ratio of the areas is $\frac{3}{4}$ , and with a infinite series, we get that the summation of the total area is:

$\frac{a}{1-q} = \frac{\frac{9\sqrt{3}}{2}}{1-\frac{3}{4}} = \frac{36\sqrt{3}}{2} = 18\sqrt{3}$

The final step is to square this, and we get our desired result, which is $(18\sqrt{3})^2 = 972$

~akliu

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2021 WSMO Team Round/Problem 4